Does the actual frequency of the sound change?

No, the source is still producing waves at the exact same rate. Only the *perceived* frequency changes because the relative motion alters how often those waves reach your ear.

Does the Doppler effect apply to light as well as sound?

Yes! The Doppler effect applies to all waves. For light, a source moving toward you shifts to a higher frequency (blueshift), and a source moving away shifts to a lower frequency (redshift). Astronomers use this to measure how fast galaxies are moving.

What happens if the source is moving faster than the speed of sound?

If the source moves faster than the waves it produces, the waves pile up entirely behind the source, creating a shock wave. This is what causes a sonic boom.

What causes the Doppler effect?

The Doppler effect is caused by the relative motion between the source of a wave and the person (or detector) observing it. When a source moves toward you, it "chases" the waves it just emitted, causing them to bunch up. This makes the waves hit you more frequently, which you perceive as a higher pitch or frequency.

Conversely, if the source is moving away, the waves are stretched out. They hit you less often, resulting in a lower perceived frequency. The actual frequency created by the source never changes; it is only your perspective of the wave that shifts.

An everyday analogy

Imagine you are standing in a pool and a friend is throwing tennis balls at you once every second. If your friend stands still, you catch one ball per second.

Now, imagine your friend starts walking toward you while continuing to throw a ball every second. Because they are stepping closer for each throw, the balls have less distance to travel. They will reach you more frequently than once a second. If they walk away from you, each ball has to travel a little further than the last, so they arrive less frequently. This is exactly what happens to sound or light waves.

The precise definition

In physics, the Doppler effect is defined as the change in the observed frequency of a wave when there is relative motion between the wave source and the observer.

The mathematical relationship for sound waves is given by the Doppler equation: $f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right)$ where $f'$ is the observed frequency, $f$ is the emitted frequency, $v$ is the speed of sound, $v_o$ is the velocity of the observer, and $v_s$ is the velocity of the source.

Where students slip up

The biggest trap in Doppler effect problems is choosing the correct plus or minus signs in the formula.

Remember this simple rule: motion toward each other always increases the observed frequency, so you need a larger numerator (use $+$ for $v_o$ ) or a smaller denominator (use $-$ for $v_s$ ). Motion away from each other decreases the frequency, meaning you need a smaller numerator ( $-$ ) or a larger denominator ( $+$ ).

Worked through

An ambulance is driving toward a stationary pedestrian at $30 \text{ m/s}$ . The ambulance's siren emits a frequency of $500 \text{ Hz}$ . If the speed of sound in air is $340 \text{ m/s}$ , what frequency does the pedestrian hear?

First, identify your knowns:

Speed of sound, $v = 340 \text{ m/s}$
Velocity of the observer (pedestrian), $v_o = 0 \text{ m/s}$
Velocity of the source (ambulance), $v_s = 30 \text{ m/s}$
Source frequency, $f = 500 \text{ Hz}$

Since the ambulance is moving toward the stationary observer, the frequency should increase. We use a minus sign in the denominator.

$f' = f \left( \frac{v + v_o}{v - v_s} \right)$ $f' = 500 \left( \frac{340 + 0}{340 - 30} \right)$ $f' = 500 \left( \frac{340}{310} \right)$ $f' \approx 500 \times 1.0967$ $f' \approx 548.4 \text{ Hz}$

The pedestrian hears a higher pitch of approximately $548 \text{ Hz}$ .

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Where this comes from: OpenStax Physics, Chapter 17: Sound · Khan Academy, Physics: The Doppler effect